Question

How do you integrate \[\cos \left( \ln x \right)dx\]?

Answer 1

Hint: In the given question, we have been asked to integrate the given expression. This question will be solved by parts method of integration such that,\[\int{f\left( x \right)g'\left( x \right)dx=f\left( x \right)g\left( x \right)-\int{f'\left( x \right)g\left( x \right)dx}}\].

For solving the question, first we need to substitute \[\ln x=\theta \ thus\ \Rightarrow x={{e}^{\theta }}\]and then differentiate it with respect to ‘x’. Later we substitute these values in the given integral and integrate the resultant expression by using the by-parts method. We will need to simplify the resultant integral to get the required integration.

Complete step-by-step solution:
We have given that,
\[\Rightarrow \int{\cos \left( \ln x \right)dx}\]
Now,
Let
\[\ln x=\theta \ thus\ \Rightarrow x={{e}^{\theta }}\]
Differentiating both the sides, we get
\[\ \Rightarrow dx={{e}^{\theta }}d\theta \]
Therefore,
Substituting the values from the above in the given integral, we get
\[\Rightarrow \int{\cos \left( \ln x \right)dx}=\int{\cos \theta \left( {{e}^{\theta }} \right)d\theta }\]
Rewrite the above as integral as,
\[\Rightarrow \int{\cos \theta \left( {{e}^{\theta }} \right)d\theta }=\int{\left( {{e}^{\theta }} \right)\left( \cos \theta \right)d\theta }\]
Let A be the integral, thus we obtained
\[\Rightarrow A=\int{\left( {{e}^{\theta }} \right)\left( \cos \theta \right)d\theta }\]
Formula of integration by parts as follows;
\[\int{f\left( x \right)g'\left( x \right)dx=f\left( x \right)g\left( x \right)-\int{f'\left( x \right)g\left( x \right)dx}}\]
Thus, integrating the resultant expression, we obtain
Here,
\[f\left( x \right)={{e}^{\theta }}\ thus\ \Rightarrow f'\left( x \right)={{e}^{\theta }}\]
And
\[g'\left( x \right)=\cos \theta \ thus\ \Rightarrow g\left( x \right)=\sin \theta \]
Therefore,
\[\Rightarrow A=\int{\left( {{e}^{\theta }} \right)\left( \cos \theta \right)d\theta }=\left( {{e}^{\theta }} \right)\left( \sin \theta \right)-\int{\left( {{e}^{\theta }} \right)\left( \sin \theta \right)d\theta }\]
Integrate again by using the by-parts method, we get
\[\Rightarrow A=\left( {{e}^{\theta }} \right)\left( \sin \theta \right)+\left( {{e}^{\theta }} \right)\left( \cos \theta \right)-\int{\left( {{e}^{\theta }} \right)\left( \cos \theta \right)d\theta }\]
Substituting \[A=\int{\left( {{e}^{\theta }} \right)\left( \cos \theta \right)d\theta }\] in the above integration, we get
\[\Rightarrow A=\left( {{e}^{\theta }} \right)\left( \sin \theta \right)+\left( {{e}^{\theta }} \right)\left( \cos \theta \right)-A\]
Simplifying the above, we get
\[\Rightarrow 2A=\left( {{e}^{\theta }} \right)\left( \sin \theta \right)+\left( {{e}^{\theta }} \right)\left( \cos \theta \right)\]
\[\Rightarrow A=\dfrac{1}{2}\left( \left( {{e}^{\theta }} \right)\left( \sin \theta \right)+\left( {{e}^{\theta }} \right)\left( \cos \theta \right) \right)=\dfrac{1}{2}{{e}^{\theta }}\left( \sin \theta +\cos \theta \right)+C\]
Thus,
\[\Rightarrow A=\dfrac{1}{2}{{e}^{\theta }}\left( \sin \theta +\cos \theta \right)+C\]
Therefore,
\[\Rightarrow \int{\cos \left( \ln x \right)dx}=\dfrac{1}{2}{{e}^{\theta }}\left( \sin \theta +\cos \theta \right)+C\]
Undo the substitution\[x=\theta \],
\[\Rightarrow \int{\cos \left( \ln x \right)dx}=\dfrac{1}{2}{{e}^{x}}\left( \sin x+\cos x \right)+C\]
Hence, this is the required integration.

Hence the correct answer is \[ \int{\cos \left( \ln x \right)dx}=\dfrac{1}{2}{{e}^{x}}\left( \sin x+\cos x \right)+C\]

Note: While solving these types of questions, students always remember the formula for integration using the by-parts method. Students mostly make mistakes while choosing the f(x) and g(x), so you should be very careful while choosing the f(x) and g(x) as it will make the given integration easy to solve easily. We should be well aware about the by-parts method of integration and we should also be well known about the application of integration. Students should be very careful while doing the calculation part of the given integral.